1 + 2 + 3 + .. +97 + 98 +99 +100 

= ( 1 + 100 ) + ( 2+ 99) + ( 3 +98) + .... + (50 + 51) 

= 101 + 101 + 101 +... +101 

Số số hạng của tổng là: 

         ( 100 - 1) : 1 + 1 = 100 ( số hạng)

Số số 101 trong tổng là: 100 : 2 = 50 số hạng 

Tổng là:

    = 101 .50 = 5050

có hai cách mà mình giải cách thứ hai thôi nhé

1+2+3+....+97+98+99+100

=(1+100)+(2+99)+(3+97)+......+(50+51)

=101+101+101+101+....+101    

mà có:100:2=50(cặp)

=101x50

=5050

dễ quá mà **** cho mình với

1 + 2 +3 + .... + 97 + 98 + 99 + 100

= ( 100 + 1 ) . 100 : 2

=  101 . 100 : 2

=    10100 : 2

=     5050

Dấu . là nhân nhé

giải luôn nha

Đặt \(A=1-3+3^2-3^3+...-3^{99}+3^{100}\)

\(\Rightarrow3A=3-3^2+3^3-...-3^{100}+3^{101}\)

\(\Rightarrow3A+A=3-3^2+3^3-...-3^{100}+3^{101}+1-3+3^2-3^3+...-3^{99}+3^{100}\)

\(\Rightarrow4A=1+3^{101}\)

\(\Rightarrow A=\dfrac{1+3^{101}}{4}\)

Ta có: \(A=1-3+3^2-3^3+...-3^{99}+3^{100}\)

\(\Leftrightarrow3\cdot A=3-3^2+3^3-3^4+...-3^{100}+3^{101}\)

\(\Leftrightarrow4\cdot A=3^{101}+1\)

hay \(A=\dfrac{3^{101}+1}{4}\)

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{203}{3^{100}}< 3\)

\(A< \frac{3}{4}\left(đpcm\right)\)

CMR: \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{100}{4^{100}}< \frac{4}{9}\)

Dạng tổng quát: CMR: \(\frac{1}{k}+\frac{2}{k^2}+\frac{3}{k^3}+\frac{4}{k^4}+...+\frac{n}{k^n}< \frac{k}{\left(k-1\right)^2}\)(k;n \(\in\) N*; k > 1)

32,48 : 0,25 + 67,52 x 4

= ( 32,48 + 67,52) : ( 0,25 x4)

= 100 : 1

= 100

1,5 được rồi bạn ạ chứ câu b mik không biết làm:)

cảm ơn bạn

đặt A=100^10+1/100^10-1

B=10^100+1/10^100-3

ta có:\(A=\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{10}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{100^{10}-1}\)

\(B=\frac{10^{100}+1}{10^{100}-3}=\frac{10^{100}-3+4}{10^{100}-3}=\frac{10^{100}-3}{10^{100}-3}+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}=1+\frac{4}{100^{10}-3}\)

vì 100¹⁰-1>100¹⁰-3

=>\(\frac{4}{100^{10}-1}<\frac{4}{100^{10}-3}\)

=>A<B

 Arcobaleno sai

Ta có:

\(\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{10}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{10^{20}-1}=1+\frac{4}{2.10^{20}-2}\)

\(\frac{10^{100}+1}{10^{100}-3}=\frac{10^{100}-3+4}{10^{100}-3}=\frac{10^{100}-3}{10^{100}-3}+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}\)

Thấy: \(2.10^{20}-2<10^{100}-3\)

\(\Rightarrow\frac{4}{2.10^{20}-1}>\frac{4}{10^{100}-3}\)

\(\Rightarrow\frac{100^{10}+1}{100^{10}-1}>\frac{10^{100}+1}{10^{100}-3}\)